cp's OEIS Frontend

Two times Chebyshev polynomials of the first kind evaluated at 3.

Also 2(a(2*n)-2) and a(2*n+1)-2 are perfect squares. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003

Chebyshev polynomials of the first kind evaluated at 3, then multiplied by 2. - Michael Somos, Apr 07 2003

Also gives solutions > 2 to the equation x^2 - 3 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004

Output of Lu and Wu's formula for the number of perfect matchings of an m X n Klein bottle where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009

It appears that for prime P = 8*n +- 3, that a((P-1)/2) == -6 (mod P) and for all composites C = 8*n +- 3, there is at least one i < (C-1)/2 such that a(i) == -6 (mod P). Only a few of the primes P of the form 8*n +-3, e.g., 29, had such an i less than (P-1)/2. As for primes P = 8*n +- 1, it seems that the sum of the two adjacent terms, a((P-1)/2) and a((P+1)/2), is congruent to 8 (mod P). - Kenneth J Ramsey, Feb 14 2012 and Mar 05 2012

For n >= 1, a(n) is also the curvature of circles (rounded to the nearest integer) successively inscribed toward angle 90 degree of tangent lines, starting with a unit circle. The expansion factor is 5.828427... or 1/(3 - 2*sqrt(2)), which is also 3 + 2*sqrt(2) or A156035. See illustration in links. - Kival Ngaokrajang, Sep 04 2013

Except for the first term, positive values of x (or y) satisfying x^2 - 6*x*y + y^2 + 32 = 0. - Colin Barker, Feb 08 2014

Albert Beiler states (page 228 of Recreations in the Theory of Numbers): D. H. Lehmer modified Lucas's test to the relatively simple form: If and only if 2^n-1 divides a(n-2) then 2^n-1 is a prime, otherwise it is composite. Since 2^3 - 1 is a factor of a(1) = 14, 2^3 - 1 = 7 is a prime. - Gary W. Adamson, Jun 07 2003

a(n) - a(n-1) divides a(n+1) - a(n). - Thomas Ordowski, Dec 24 2016

Indices m of triangular numbers T(m) which are one-third of another triangular number: 3*T(m) = T(k); the k's are given by A001571. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002

On the previous comment: for m=0 this is actually one third of the same triangular number. - Zak Seidov, Apr 07 2011

Also numbers n such that the n-th centered 24-gonal number 12*n*(n+1)+1 is a perfect square A001834(n)^2, where A001834(n) is defined by the recursion: a(0) = 1, a(1) = 5, a(n) = 4*a(n-1) - a(n-2) + 1. - Alexander Adamchuk, Apr 21 2007

Also numbers n such that RootMeanSquare(5,...,6*n-1) is an integer. - Ctibor O. Zizka, Dec 17 2008 (Corrected by Robert K. Moniot, Jul 22 2020)

Also numbers n such that n*(n+1) = Sum_{i=1..x} n+i for some x. (This does not apply to the first term.). - Gil Broussard, Dec 23 2008

From John P. McSorley, May 26 2020: (Start)

Consecutive terms (a(n-1), a(n)) = (u,v) give all points on the hyperbola u^2 - u + v^2 - v - 4*u*v = 0 in quadrant I with both coordinates an integer.

Also related to the block sizes of small multi-set designs. (End)

If a(n) white balls and a(n+1) black balls are mixed in a bag, and a pair of balls is drawn without replacement, the probability that one ball of each color is drawn is exactly 1/3. These are the only integers for which the probability is 1/3. For example, if there are 20 white balls and 76 black balls, the probability of drawing one of each is (20/96)*(76/95) + (76/96)*(20/95) = 1/3. - Elliott Line, May 13 2022

Numerators of continued fraction convergents to sqrt(3), for n >= 1.

For the denominators see A002530.

Consider the mapping f(a/b) = (a + 3*b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the convergents 1/1, 2/1, 5/3, 7/4, 19/11, ... converging to sqrt(3). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003

In the Murthy comment if we take a = 0, b = 1 then the denominator of the reduced fraction is a(n+1). A083336(n)/a(n+1) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 26 2003

If signs are disregarded, all terms of A002316 appear to be elements of this sequence. - Creighton Dement, Jun 11 2007

2^(-floor(n/2))*(1 + sqrt(3))^n = a(n) + A002530(n)*sqrt(3); integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 10 2018

Let T(n) = A000034(n), U(n) = A002530(n), V(n) = a(n), x(n) = U(n)/V(n). Then T(n*m) * U(n+m) = U(n)*V(m) + U(m)*V(n), T(n*m) * V(n+m) = 3*U(n)*U(m) + V(m)*V(n), x(n+m) = (x(n) + x(m))/(1 + 3*x(n)*x(m)). - Michael Somos, Nov 29 2022

Least side in a triangle with integer sides (m, m+1, m+2) (m >= 1) and integer area. The degenerate triangle with sides (1,2,3) is included.

Also describes triangles whose sides are consecutive integers and in which the inscribed circle has an integer radius. - Harvey P. Dale, Dec 28 2000 [Then, the length of this inradius is A001353(n). - Bernard Schott, Mar 21 2023]

Equivalently, positive integers m such that (3/16)*m^4 + (3/4)*m^3 + (3/8)*m^2 - (3/4)*m - 9/16 is a square (A000290), a direct result of Heron's formula. - Rick L. Shepherd, Sep 04 2005

"The problem is to find the sides of a triangle that shall have the values n, n + 1, and n + 2 and such that the perpendicular upon the longest side from the opposite vertex shall be rational. Nakane solves it as follows..." (Smith and Mikami, 2004). - Jonathan Sondow, May 09 2013

For n >= 1 all terms are congruent to {1,3} mod 10. Among first 100 terms there are 6 prime numbers: 3, 13, 193, 37633, 7300801, 1416317953. - Zak Seidov, Jun 14 2018

n > 1 is in this sequence if and only if the triangle with sides 4, n, n+2 has integer area (compare with A072221 for sides 3, n, n+1). - Michael Somos, May 11 2019

a(0) = 1 corresponds to the degenerate triangle [1,2,3], with area = 0. - Wesley Ivan Hurt, May 20 2020

Since this is a list it should really have offset 1, but that would require a large number of changes. - N. J. A. Sloane, Feb 04 2021

Least distance from centroid of a triangle to vertices, distances being m, m+1, m+2 and triangle area being a nonnegative integer. - Alexandru Petrescu, Feb 28 2023

Then, in this case, with a(n) = m, the corresponding area of this triangle is 3 * A011945(n+1). - Bernard Schott, Mar 21 2023

For n > 1, a(n-1) is the determinant of the n X n band matrix which has {2,4,4,...,4,4,2} on the diagonal and a 1 on the entire super- and subdiagonal. This matrix appears when constructing a natural cubic spline interpolating n equally spaced data points. - g.degroot(AT)phys.uu.nl, Feb 14 2007

Integer values of x that make 9+3*x^2 a perfect square. - Lorenz H. Menke, Jr., Mar 26 2008

The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence whose numerators are the terms of this sequence and denominators are A001075. - Clark Kimberling, Aug 27 2008

a(n) also give the altitude to the middle side of a Super-Heronian Triangle. - Johannes Boot, Oct 14 2010

a(n) gives values of y satisfying 3*x^2 - 4*y^2 = 12; corresponding x values are given by A003500. - Sture Sjöstedt, Dec 19 2017

Corresponding m's are in A016064. Corresponding values of lesser side give A016064.

a(n) gives the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6, 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a rook on the four side and four corner (m = 1, 3, 7, 9) squares but on the center square (m = 5) it goes berserk and turns into a berserker. For this sequence, the berserker can move to three of the side squares and three of the corners from the center.

The berserker is one of the Lewis chessmen which were discovered in 1831 on the Isle of Lewis. They are carved from walrus ivory in Scandinavian style of the 12th century. The pawns look like decorated tombstones. The pieces have all human representations with facial expressions varying from gloom to anger. Some of the rooks show men biting their shield in the manner of berserkers. According to Hooper and Whyld none looks happy.

Let A be the adjacency matrix of the graph G, where V(G) = {v1, v2, v3, v4, v5, v6, v7, v8, v9}. Then the (m, k) entry of A^n is the number of different vm-vk walks of length n in G, see the Chartrand reference. In the adjacency matrix A, see the Maple program, the A[1], A[3], A[7] and A[9] vectors represent the rook moves on the corner squares, the A[2], A[4], A[6] and A[8] vectors represent the rook moves on the side squares and the A[5] vector represents the moves of the berserker. On a 3 X 3 chessboard there are 2^9 = 512 ways a berserker could move from the center square (off the center the berserker behaves like a rook) so there are 512 different berserkers.

For the side squares the 512 berserker vectors lead to 42 different sequences, see the overview of berserker sequences. There are 16 berserker vectors that lead to the sequence given above. Their decimal [binary] values are: 111 [001 101 111] , 207 [011 001 111], 231 [011 100 111], 237 [011 101 101], 303 [100 101 111], 363 [101 101 011], 366 [101 101 110], 399 [110 001 111], 423 [110 100 111], 429 [110 101 101], 459 [111 001 011], 462 [111 001 110], 483 [111 100 011], 486 [111 100 110], 489 [111 101 001] and 492 [111 101 100]. These berserker vectors lead for the corner squares to sequence 4*A179606 (with leading term 1 added) and for the central square to sequence 6*A179606 (with leading term 1 added).

This sequence belongs to a family of sequences with GF(x)=(1+x-k*x^2)/(1-3*x+(k-4)*x^2), see A180142.

Positive integers k such that x^2 - 4xy + y^2 + k = 0 has integer solutions. (See the CROSSREFS section for sequences relating to solutions for particular k.)

Comments on method used, from Colin Barker, Jun 06 2014: (Start)

In general, we want to find the values of f, from 1 to 400 say, for which x^2 + bxy + y^2 + f = 0 has integer solutions for a given b.

In order to solve x^2 + bxy + y^2 + f = 0 we can solve the Pellian equation x^2 - Dy^2 = N, where D = b*b - 4 and N = 4*(b*b - 4)*f.

But since sqrt(D) < N, the classical method of solving x^2 - Dy^2 = N does not work. So I implemented the method described in the 1998 sci.math reference, which says:

"There are several methods for solving the Pellian equation when |N| > sqrt(d). One is to use a brute-force search. If N < 0 then search on y = sqrt(abs(n/d)) to sqrt((abs(n)(x1 + 1))/(2d)) and if N > 0 search on y = 0 to sqrt((n(x1 - 1))/(2d)) where (x1, y1) is the minimum positive solution (x, y) to x^2 - dy^2 = 1. If N < 0, for each positive (x, y) found by the search, also take (-x, y). If N > 0, also take (x, -y). In either case, all positive solutions are generated from these using (x1, y1) in the standard way."

Incidentally all my Pell code is written in B-Prolog, and is somewhat voluminous. (End)

Also, positive integers of the form -x^2 + 2xy + 2y^2 of discriminant 12. - N. J. A. Sloane, May 31 2014 [Corrected by Klaus Purath, May 07 2023]

The equivalent sequence for x^2 - 3xy + y^2 + k = 0 is A031363.

The equivalent sequence for x^2 - 5xy + y^2 + k = 0 is A237351.

A positive k does not appear in this sequence if and only if there is no integer solution of x^2 - 3*y^2 = -k with (i) 0 < y^2 <= k/2 and (ii) 0 <= x^2 <= k/2. See the Nagell reference Theorems 108 a and 109, pp. 206-7, with D = 3, N = k and (x_1,y_1) = (2,1). - Wolfdieter Lang, Jan 09 2015

From Klaus Purath, May 07 2023: (Start)

There are no squares in this sequence. Products of an odd number of terms as well as products of an odd number of terms and any terms of A014209 belong to the sequence.

Products of an even number of terms are terms of A014209. The union of this sequence and A014209 is closed under multiplication.

A positive number belongs to this sequence if and only if it contains an odd number of prime factors congruent to {2, 3, 11} modulo 12. If it contains prime factors congruent to {5, 7} modulo 12, these occur only with even exponents. (End)

From Klaus Purath, Jul 09 2023: (Start)

Any term of the sequence raised to an odd power also belongs to the sequence. Proof: t^(2n+1) = t*t^2n = (3*x^2 - y^2)*t^2n = 3*(x*t^n)^2 - (y*t^n)^2. It seems that t^(2n+1) occurs only if t also is in the sequence.

Joerg Arndt has proved that there are no squares in this sequence: Assume s^2 = 3*y^2 - x^2, then s^2 + x^2 = 3 * y^2, but the sum of two squares cannot be 3 * y^2, qed. (End)

That products of any 3 terms belong to the sequence can be proved by the following identity: (na^2 - b^2) (nc^2 - d^2) (ne^2 - f^2) = n[a(nce + df) + b(cf + de)]^2 - [na(cf + de) + b(nce + df)]^2. This can be verified by expanding both sides of the equation. - Klaus Purath, Jul 14 2023

y-values in the solutions to 3*x^2+6 = y^2. - Sture Sjöstedt, Nov 25 2011

Positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 18 = 0. - Colin Barker, Feb 04 2014

Positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 288 = 0. - Colin Barker, Feb 16 2014